prove that following set is closed in $\mathbb{R}^n$

Question

Can someone verify proof of following statement.

prove that $\hat{B}(\textbf{a} , \delta) = \{\textbf{x} \in \mathbb{R}^n : ||\textbf{x} - \textbf{a}|| \leq \delta \}$ is closed

1. let $\{\textbf{x}_{k}\} \rightarrow \textbf{x}$ be some convergent sequence that is completely in $\hat{B}(\textbf{a} ,\delta)$

2. we have to show that $\textbf{x} \in \hat{B}(\textbf{a} , \delta)$ that is $||\textbf{x} - \textbf{a}|| \leq \delta$

3. since we know that $\{\textbf{x}_k\}$ is in $\hat{B}(a,\delta)$ we have that $||\textbf{x}_k - \textbf{a}|| \leq \delta$ forall $k \in \mathbb{N}$

4. also, as we have that $\lim_{k \rightarrow \infty} \textbf{x}_k = \textbf{x}$

5. there exists $N \in \mathbb{N} \,\, \forall k \geq N ||\textbf{x}_k - \textbf{x}|| < \epsilon$

6. $||\textbf{x} - \textbf{a}|| = ||\textbf{x} - \textbf{x}_{k} + \textbf{x}_k - \textbf{a} || \leq || \textbf{x} - \textbf{x}_k || + ||\textbf{x}_k - \textbf{a}|| \leq \epsilon + \delta$

7. since this holds for arbitrary $\epsilon$ we have that $||\textbf{x} - \textbf{a}|| \leq \delta$

8. we are done

Answer 1

Your proof is O.K. Between 4. and 5. you should spend " let $\varepsilon >0$ be given".

Answer 2

I think you're way overthinking this: if $\mathbf{x} \notin \hat{B}(\mathbf{a}, \delta)$, then $\|\mathbf{x}-\mathbf{a}\| > \delta$.

Then show directly with the triangle inequality that $B(\mathbf{x},r) \cap \hat{B}(\mathbf{a}, \delta) = \emptyset$ for $r:= (\|\mathbf{x}-\mathbf{a}\| - \delta) >0$. Or put otherwise: $B(\mathbf{x},r) \subseteq \Bbb R^n\setminus \hat{B}(\mathbf{a}, \delta)$, so the complement is open.

That proof works in any metric space, no fact about $\Bbb R^n$ is needed beyond that, and no sequences are needed either.